Organic Chemistry Practice Problems and Problem Sets
alkene addition
Let's work through a halogenation reaction. Draw the structures for each of the species in the four boxes below (3º carbocation, halonium ion, protonated thiol, and thiol). Also draw curved arrows to show electron movement. Note: thiol = RSH, like an alcohol, but with sulfur instead of oxygen.
Because the chlorine is more electronegative than iodine, the iodine will have a partial positive charge and will be attacked by the alkene. It forms the more stable carbocation as normal (as in problem 335). The nucleophile (HSCH2CH3) will eventually attack the carbocation, but the iodine does something special first- it forms a cyclic intermediate.
The formation of the cyclic halonium ion is probably the most important concept you will learn in first semester organic chemistry. With iodine, it's called an iodonium ion. (with chlorine or bromine, it's called a chloronium or bromonium ion).
The HSCH2CH3 will attack the carbon that was the carbocation, but it's forced to attack from the side opposite to that of the iodine. So the product will always have anti stereochemistry (you get the trans product).
MendelSet practice problem # 521 submitted by Matt on July 1, 2011.
Let's work through an alkene addition reaction. Draw the structures for each of the species in the three boxes below (3º carbocation, protonated thiol, and thiol). Also draw curved arrows to show electron movement. Note: thiol = RSH, like an alcohol, but with sulfur instead of oxygen.
Note that this is the reverse of problem 519. Instead of going from thiol (alcohol) to carbocation to alkene, we're going from alkene to carbocation to thiol. In each step ask yourself "what arrows can I draw?" and choose the step that doesn't go backwards.
The first step is the alkene picks up a proton to form the more stable carbocation. (See problem 334 if you don't understand why only the 3º carbocation is formed). To get rid of the carbocation, we can either do a beta-elimination (E1) to form an alkene, or an addition reaction (SN1) to form a protonated thiol. Since forming the alkene would be going backwards, the only choice is addition.
Then you're left with a protonated thiol. How do we get rid of the positive charge on the sulfur? There are two legal moves- either the RSCH2CH3 acts as a leaving group, or the sulfur deprotonates. Since the RSCH2CH3 leaving would form a carbocation and take us backwards, the only option is for the sulfur to deprotonate.
Because the sulfur added to the more substituted carbon, this was a Markovnikov addition.
MendelSet practice problem # 520 submitted by Matt on June 30, 2011.
For the reaction below, draw the structures of the radical intermediate and the final product.
When HBr reacts with in an alkene in the presence of peroxides (which are free radical initiators), the reaction goes through a free radical mechanism and adds in an anti-Markovnikov manner. The product is an alkyl bromide.
The reaction goes anti-Markovnikov because the more substituted radical intermediate is only formed if the bromine adds to the less substituted carbon.
MendelSet practice problem # 345 submitted by Matt on June 7, 2011.
For the reaction below, draw the structures of the borane intermediate and the final product.
This is a hydroboration reaction (also called hydroboration-oxidation in some textbooks). The product is an alcohol.
Boron bonds to the less substituted carbon, and is replaced with an oxygen after the borane intermediate is treated with peroxide, resulting in an anti-Markovnikov addition.
Note that one equivalent of BH3 reacts with three equivalents of alkene; the borane intermediate is not BH2(alkyl) but rather B(alkyl)3.
MendelSet practice problem # 344 submitted by Matt on June 7, 2011.